The Notebook
İstanbul, Turkey

# Proof on Infinity of Prime Numbers

In this post, I show that we have infinitely many prime numbers. To do so, I use a proof from 300 BC which provided by Greek Mathematician. The one that people usually call as “Father of Geometry” or Euclid of Alexandria.

Here we break into proof:

Let's write prime numbers in an ascending manner, p_1 = 2, p_2 = 3, p_3 = 5, p_4 = 7

When we check the numbers which are in a pattern of: p_1.p_2 + 1 = 2.3 + 1 = 7 and p_1.p_2.p_3 + 1 = 2.3.5 + 1 = 31 and p_1.p_2.p_3.p_4 + 1 = 2.3.5.7 + 1 = 211 we oftenly come up with a prime number. But how can we be sure about primality of number A = p_1.p_2 \dots p_n + 1 for all values of n?

Here we use contradiction technique to check: We intuitively think that A is a prime number, since it empirically is. Now we set up an equation which assumes that A is a non-prime number and try to show that this equation would not hold -hence contradicts.

Following equation assumes that A is a divisor of some number. We clearly know that, every number -except 1- is at least multiple of one prime number. For this reason, let's say that p_r is a divisor of A.

A = p_1.p_2.p_3 \dots p_r \dots p_n + 1 \rightarrow \frac{A}{p_r} = p_1.p_2.p_3 \dots p_{r-1}.p_{r+1} \dots p_n + \frac{1}{p_r}. Consequently we have \frac{A}{p_r} - p_1.p_2.p_3 \dots p_{r-1}.p_{r+1} \dots p_n = \frac{1}{p_r}

Let's concentrate on last equation. \frac{A}{p_r} has to be an integer, since we assume A is a multiple of p_r. Second term p_1.p_2.p_3 \dots p_{r-1}.p_{r+1} \dots p_n is also integer, nevertheless \frac{1}{p_r} is not. So, our non-primality assumption of A was wrong and this wrong assumption proves us that, A would always be prime for any value of n.